TITLE: Does existence of anti-derivative imply integrability?
QUESTION [17 upvotes]: If $f$ has an anti-derivative in $[a,b]$ does it imply that $f$ is Riemann integrable in $[a,b]$?

REPLY [17 votes]: Take $f(x)=\begin{cases} x^2\sin (1/x^2), &x\ne 0, \\ 0, &x=0. \end{cases}\quad$ Then $g=f'$ exists everywhere but is unbounded over $[-1,1]$. $g$ thus has a primitive but is not Riemann integrable.